All you need to do is find $\displaystyle $$\theta_{0.9}$ such that:
$\displaystyle P(X\ge x|\theta_{0.9})<0.1$
where $\displaystyle $$x$ is the observation, which is the solution for $\displaystyle $$\theta_{0.9}$ to:
$\displaystyle e^{-x/\theta_{0.9}}=0.9$
which is:
$\displaystyle \displaystyle \theta_{0.9}=\frac{x}{\ln(0.9)}$
CB